5.7 Forced Mechanical Vibrations
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5.7 Forced Mechanical Vibrations
The study of vibrating mechanical systems continues. The main example is a system consisting of an externally forced mass on a spring with
dampener. Both undamped and damped systems are studied. A number of physical examples are given, which include the following: clothes
dryer, cafe door, pet door, bicycle trailer.
Forced undamped motion
The equation for study is a forced spring–mass system
mx′′ (t) + kx(t) = f (t).
The model originates by equating the Newton’s second law force mx′′ (t)
to the sum of the Hooke’s force −kx(t) and the external force f (t).
The physical model is a laboratory box containing an undamped spring–
mass system, transported on a truck as in Figure 11, with external force
f (t) = F0 cos ωt induced by the speed bumps.
k
x=0
x>0
Figure 11. An undamped
spring-mass system in a box is
transported on a truck. Speed
bumps on the shoulder of the
road induce periodic vertical
oscillations to the box.
m
The forced equation takes the form
x′′ (t) + ω02 x(t) =
F0
cos ωt,
m
ω0 =
q
k/m.
The natural frequency ω0 corresponds to free oscillation of the mass,
that is, the number of full periods of oscillation per second for the spring–
mass system when no external force is present. The external frequency
ω is the number of full periods of oscillation per second of the external
force f (t) = F0 cos ωt. In the case of Figure 11, this is the vertical force
applied to the box containing the spring–mass system, due to the speed
bumps. The general solution x(t) always presents itself in two pieces, as
the sum of the homogeneous solution xh and a particular solution xp .
For ω 6= ω0 , the general solution is
(1)
x(t) = xh (t) + xp (t),
xh (t) = c1 cos ω0 t + c2 sin ω0 t, c1 , c2 constants,
F0 /m
.
xp (t) = F1 cos ωt, F1 = 2
ω0 − ω 2
A general statement can be made about the solution decomposition:
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The solution is a sum of two harmonic oscillations, one of
natural frequency ω0 due to the spring and the other of
natural frequency ω due to the external force F0 cos ωt.
Rapidly and slowly varying functions. The superposition x(t)
will show the phenomenon of beats for certain choices of ω0 , ω, x(0)
and x′ (0). For example, consider x(t) = cos ω0 t − cos ωt. Use the
trigonometric identity 2 sin a sin b = cos(a − b) − cos(a + b) to write
x(t) = A(t) sin 21 (ω0 + ω)t where A(t) = 2 sin 21 (ω0 − ω)t. If ω ≈ ω0 ,
then A(t) has natural frequency α = 12 (ω0 − ω) near zero. The natural
frequency β = 12 (ω0 + ω) can be relatively large and therefore x(t) is a
product of a slowly varying amplitude A(t) = 2 sin αt and a rapidly
varying oscillation sin βt.
The physical phenomenon of beats refers to the periodic cancellation of
sound at a slow frequency. An illustration of the graphical meaning of
beats appears in Figure 12.
x
t
Figure 12. The phenomenon of
beats. Shown is a rapidly–varying
periodic oscillation x(t) = 2 sin 4t sin 40t
and the two slowly–varying envelope
curves x1 (t) = 2 sin 4t, x2 (t) = −2 sin 4t.
An example is striking simultaneously two identical tuning forks, the
first slightly out of tune with the second. A destructive interference
occurs during a very brief interval, so our impression is that the sound
periodically stops, only briefly, and then starts again with a beat, a section that is instantaneously loud again. The origin of this impression
can be seen from the formula x(t) = A(t) sin βt where A(t) = 2 sin αt.
There is no sound when x(t) ≈ 0: this is when destructive interference
occurs. When α is small compared to β, there are long intervals between
the zeros of A(t), at which destructive interference occurs and x(t) ≈ 0.
Otherwise, the amplitude of the sound is the average value of A(t), which
is 1. The sound stops at a zero of A(t) and then it is rapidly loud again,
causing the beat.
Rotating drum on a cart. Figure 13 shows a model for a rotating
machine, like a front–loading clothes dryer.
For modelling purposes, the rotating drum with load is replaced by an
idealized model: a mass M on a string of radius R rotating with angular
speed ω. The center of rotation is located along the center–line of the
cart. The total mass m of the cart includes the rotating mass M, which
we imagine to be an off–center lump of wet laundry inside the dryer
drum. Vibrations cause the cart to skid left or right.
5.7 Forced Mechanical Vibrations
k
R θ
0 x
225
M
Figure 13. A rotating vertical drum
installed on a cart with skids. A
spring restores the cart to its
equilibrium position x = 0.
A spring of Hooke’s constant k restores the cart to its equilibrium position x = 0. The cart has position x > 0 corresponding to skidding
distance x to the right of the equilibrium position, due to the off-center
load. Similarly, x < 0 means the cart skidded distance |x| to the left.
Modelling. Friction ignored, Newton’s second law gives force F =
mx′′ (t), where x locates the cart’s center of mass. Hooke’s law gives
force F = −kx(t). The centroid x can be expanded in terms of x(t) by
using calculus moment of inertia formulas. Let m1 = m − M be the cart
mass, m2 = M the drum mass, x1 = x(t) the moment arm for m1 and
x2 = x(t) + R cos θ the moment arm for m2 . Then θ = ωt in Figure 13
gives
m 1 x1 + m 2 x2
x(t) =
m1 + m2
(m − M)x(t) + M(x(t) + R cos θ)
(2)
=
m
RM
= x(t) +
cos ωt.
m
Force competition mx′′ = −kx and derivative expansion results in the
forced harmonic oscillator
(3)
mx′′ (t) + kx(t) = RMω 2 cos ωt.
Forced damped motion
Real systems do not exhibit idealized harmonic motion, because damping occurs. A watch balance wheel submerged in oil is a key example:
frictional forces due to the viscosity of the oil will cause the wheel to
stop after a short time. The same wheel submerged in air will appear to
display harmonic motion, but indeed there is friction present, however
small, which slows the motion.
Consider a spring–mass system consisting of a mass m and a spring with
Hooke’s constant k, with an added dashpot or dampener, depicted in
Figure 14 as a piston inside a cylinder attached to the mass. A useful
physical model, for purposes of intuition, is a screen door with door–
closer: the closer has a spring and an adjustable piston–cylinder style
dampener.
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k
m
c
Figure 14. A spring-mass system with
dampener
The dampener is assumed to operate in the viscous domain, which
means that the force due to the dampener device is proportional to the
speed that the mass is moving: F = cx′ (t). The number c ≥ 0 is
called the damping constant. Three forces act: (1) Newton’s second
law F1 = mx′′ (t), (2) viscous damping F2 = cx′ (t) and (3) the spring
restoring force F3 = kx(t). The sum of the forces F1 + F2 + F3 acting on
the system must equal the external force f (t), which gives the equation
for a damped spring–mass system
mx′′ (t) + cx′ (t) + kx(t) = f (t).
(4)
The motion is called damped if c > 0 and undamped if c = 0. If there
is no external force, f (t) = 0, then the motion is called free or unforced
and otherwise it is called forced.
A useful visualization for a forced system is a vertical laboratory spring–
mass system with dampener placed inside a box, which is transported
down a washboard road inside an auto trunk. The function f (t) is the
vertical oscillation of the auto trunk. The function x(t) is the motion of
the mass in response to the washboard road. See Figure 15.
k
x=0
x>0
m
m
Figure 15. A spring-mass system
with dampener in a box transported
in an auto trunk along a washboard
road.
Free damped motion. Consider the special case of no external
force, f (t) = 0. The motion x(t) satisfies the homogeneous differential
equation
(5)
mx′′ (t) + cx′ (t) + kx(t) = 0.
Cafe door. Restaurant waiters and waitresses are used to the cafe
door, which blocks partially the view of onlookers, but allows rapid,
collision-free trips to the kitchen – see Figure 16. The door is equipped
with a spring which tries to restore the door to the equilibrium position
x = 0, which is the plane of the door frame. There is a dampener
attached, to keep the number of oscillations low.
5.7 Forced Mechanical Vibrations
227
Figure 16. A cafe door on three hinges with
dampener in the lower hinge. The equilibrium
position is the plane of the door frame.
The top view of the door, Figure 17, shows how the angle x(t) from
equilibrium x = 0 is measured from different door positions.
x<0
x=0
x>0
Figure 17. Top view of a cafe door,
showing the three possible door
positions.
The figure shows that, for modelling purposes, the cafe door can be
reduced to a torsional pendulum with viscous damping. This results in
the cafe door equation
Ix′′ (t) + cx′ (t) + κx(t) = 0.
(6)
The removal of the spring (κ = 0) causes the solution x(t) to be monotonic, which is a reasonable fit to a springless cafe door.
Pet door. Designed for dogs and cats, the small door in Figure 18
allows animals to enter and exit the house freely. Winter drafts and
summer insects are the main reasons for pet doors. Owners argue that
these doors decrease damage due to clawing and beating the door to get
in and out. A pet door might have a weather seal and a security lock.
Figure 18. A pet door. The equilibrium
position is the plane of the door frame.
The pet door swings freely from hinges along the top edge. One hinge is
spring–loaded with dampener. Like the cafe door, the spring restores the
door to the equilibrium position while the dampener acts to eventually
stop the oscillations. However, there is one fundamental difference: if the
spring–dampener system is removed, then the door continues to oscillate!
The cafe door model will not describe the pet door.
For modelling purposes, the door can be compressed to a linearized
swinging rod of length L (the door height). The torque I = mL2 /3
of the door assembly becomes important, as well as the linear restoring
force kx of the spring and the viscous damping force cx′ of the dampener.
All considered, a suitable model is the pet door equation
(7)
mgL
x(t) = 0.
I x (t) + cx (t) + k +
2
′′
′
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Derivation of (7) is by equating to zero the algebraic sum of the forces.
Removing the dampener and spring (c = k = 0) gives a harmonic oscillator x′′ (t) + ω 2 x(t) = 0 with ω 2 = 0.5mgL/I, which establishes sanity for
the modelling effort. Equation (7) is formally the cafe door equation with
an added linearization term 0.5mgLx(t) obtained from 0.5mgL sin x(t).
Modelling. The cafe door and the pet door have equations in the
same form as a damped spring–mass system, and all three can be reduced, for suitable definitions of constants p and q, to the simplified
second order differential equation
(8)
x′′ (t) + p x′ (t) + q x(t) = 0.
Tuning a dampener. The pet door and the cafe door have dampeners with an adjustment screw. The screw changes the damping coefficient
c which in turn changes the size of coefficient p in (8). More damping c
means p is larger.
There is a critical damping effect for a certain screw setting: if the damping is decreased more, then the door oscillates, whereas if the damping is
increased, then the door has a monotone non-oscillatory behavior. The
monotonic behavior can result in the door opening in one direction followed by slowly settling to exactly the door jamb position. If p is too
large, then it could take 10 minutes for the door to close!
The critical case corresponds to the least p > 0 (the smallest damping
constant c > 0) required to close the door with this kind of monotonic
behavior. The same can be said about decreasing the damping: the
more p is decreased, the more the door oscillations approach those of no
dampener at all, which is a pure harmonic oscillation.
As viewed from the characteristic equation r 2 + pr + q = 0, the change
is due to a change in character of the roots from real to complex. The
physical response and the three cases of the constant–coefficient recipe,
page 189, lead to the following terminology.
Classification
Defining properties
Overdamped
Distinct real roots r1 6= r2
Positive discriminant
x = c1 er1 t + c2 r r2 t
= exponential × monotonic function
Critically damped
Double real root r1 = r2
Zero discriminant
x = c1 er1 t + c2 t r r1 t
= exponential × monotonic function
5.7 Forced Mechanical Vibrations
229
Complex conjugate roots α ± i β
Negative discriminant
x = eαt (c1 cos βt + c2 sin βt)
= exponential × harmonic oscillation
Underdamped
Bicycle trailer. An auto tows a one–wheel trailer over a washboard
road. Shown in Figure 19 is the trailer strut, which has a single coil
spring and two dampeners. The mass m includes the trailer and the
bicycles.
k
c
c
Figure 19. A trailer strut with
dampeners on a washboard road
Suppose a washboard dirt road has about 2 full oscillations (2 bumps
and 2 valleys) every 3 meters and a full oscillation has amplitude 6
centimeters. Let s denote the horizontal distance along the road and let
ω be the number of full oscillations of the roadway per unit length. The
oscillation period is 2π/ω, therefore 2π/ω = 3/2 or ω = 4π/3. A model
for the road surface is
5
cos ωs.
y=
100
Let x(t) denote the vertical elongation of the spring, measured from
equilibrium. Newton’s second law gives a force F1 = mx′′ (t) and the
viscous damping force is F2 = 2cx′ (t). The trailer elongates the spring
by x − y, therefore the Hooke’s force is F3 = k(x − y). The sum of the
forces F1 + F2 + F3 must be zero, which implies
mx′′ (t) + 2cx′ (t) + k(x(t) − y(t)) = 0.
Write s = vt where v is the speedometer reading of the car in meters
per second. The expanded differential equation is the forced damped
spring-mass system equation
k
cos(4πvt/3).
20
The solution x(t) of this model, with x(0) and x′ (0) given, describes the
vertical excursion of the trailer bed from the roadway. The observed
oscillations of the trailer are modeled by the steady-state solution
mx′′ (t) + 2cx′ (t) + kx(t) =
xss (t) = A cos(4πvt/3) + B sin(4πvt/3),
where A, B are constants determined by the method√of undetermined
coefficients. From the physical data, the amplitude A2 + B 2 of this
oscillation might be 6cm or larger.
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Exercises 5.7